#### Answer

$x=\left\{ 3,5 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given rational equation, $
\dfrac{2x-5}{x}=\dfrac{x-2}{3}
,$ use cross-multiplication and the concepts of quadratic equations.
$\bf{\text{Solution Details:}}$
Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to
\begin{array}{l}\require{cancel}
(2x-5)(3)=x(x-2)
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac$ and the properties of equality, the expression above is equivalent to
\begin{array}{l}\require{cancel}
6x-15=x^2-2x
\\\\
-x^2+(6x+2x)-15=0
\\\\
-x^2+8x-15=0
\\\\
-1(-x^2+8x-15)=(0)(-1)
\\\\
x^2-8x+15=0
.\end{array}
In the trinomial expression above the value of $ac$ is $
1(15)=15
$ and the value of $b$ is $
-8
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-3,-5
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x^2-3x-5x+15=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^2-3x)-(5x-15)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(x-3)-5(x-3)=0
.\end{array}
Factoring the $GCF=
(3x-4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x-3)(x-5)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x-3=0
\\\\\text{OR}\\\\
x-5=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-3=0
\\\\
x=3
\\\\\text{OR}\\\\
x-5=0
\\\\
x=5
.\end{array}
Upon checking, none of the solutions above will make the denominator of the original equation equal to zero. Hence, $
x=\left\{ 3,5 \right\}
.$